212 BINOMIAL COEFFICIENTS
into to but disappear from t,he term ratio. Using such tricks we can predict without further calculation t;hat the term ratio of (5.27) is k - r tk+l -=fk
k -n k+l k+s-n+l
times (--1 )’ = 1, and Vandermonde’s convolution becomes (5.91)
We can use this equation to determine F( a, b; c; z) in general, when z = 1 and when b is a negative integer. Let’s rewrite (5.91) in a form so that table lookup is easy when a new sum needs to be evaluated. The result turns out to be F
a,b , _ ( C 1)
T(c-a--b)T(c) r(c - a) T(c - b) ’
integer b 6 0 or %c >Ra+!Xb.
(5.92)
Vandermonde’s convolution (5.27) covers only the case that one of the upper parameters, say b, is a nonpositive integer; but Gauss proved that (5.92) is valid also when a, b, c are complex numbers whose real parts satisfy !Xc > %a + %b. In other cases, the infinite series F( “;” j 1) doesn’t converge. When b = -n, the identity can be written more conveniently with factorial powers instead of Gamma functions: F(a’;ni,) = k&z
= (;-;s,
integer n > 0.
(5.93)
It turns out that all five of the identities in Table 169 are special cases of Vandermonde’s convolution; formula (5.93) covers them all, when proper attention is paid to degenerate situations. Notice that (5.82) is just the special case a = 1 of (5.93). Therefore we don’t really need to remember (5.82); and we don’t really need the identity (5.9) that led us to (5.82), even though Table 174 said that it was memorable. A computer program for formula manipulation, faced with the problem of evaluating xkGn (‘+kk), could convert the sum to a hypergeometric and plug into the general identity for Vandermonde’s convolution. Problem 1 in Section 5.2 asked for the value of
This problem is a natural for hypergeometrics, and after a bit of practice any hypergeometer can read off the parameters immediately as F( 1, -m; -n; 1). Hmmm; that problem was yet another special takeoff on Vandermonde!
A few weeks ago, we were studying what ~~~r~~r~e~e jn Now we’re studying stuff beyond his Ph.D. thesis. Is this intimidating or what?
